Ncert Exercises & Solutions

Three copper blocks of masses \(M_{1} , M_{2} \) and \(M_{3}~\text{kg}\) respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at \(T_{1} , T_{2} , T_{3} \left(\right. T_{1} > T_{2} > T_{3} \left.\right).\) Assuming there is no heat loss to the surroundings, the equilibrium temperature \(T\) is:
(\(s\) is the specific heat of copper)

1. \(T = \dfrac{T_{1} + T_{2} + T_{3}}{3}\)

2. \(T = \dfrac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{M_{1} + M_{2} + M_{3}}\)

3. \(T = \dfrac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{3 \left(\right. M_{1} + M_{2} + M_{3} \left.\right)}\)

4. \(T = \dfrac{M_{1} T_{1} s + M_{2} T_{2} s + M_{3} T_{3} s}{M_{1} + M_{2} + M_{3}}\)

Hint: The heat lost by one block will be equal to the heat gained by the other two blocks.

Step 1: Find the value of heat lost by one block and the heat gained by the other two blocks.

Let the equilibrium temperature of the system is \(T.\)

Let us assume that \(𝑇 _1 , 𝑇 _2 < 𝑇 < 𝑇 _3 . \)

Since there is no heat loss to the surroundings, the net heat exchange must be zero: Heat gained+Heat lost=0
The heat exchanged by each block is given by: \(Q=msΔT\)
where \(m\) is the mass, \(s\) is the specific heat, and \(\Delta T\) is the change in temperature.
For block 1 (initially at \(T_1\)):
Heat lost \(=M_1 s (T_1 - T)~~~...(1)\)
For block 2 (initially at \(T_2\)):
Heat gained/lost \(=M_2 s (T - T_2)~~~...(2)\)
(This block can either gain or lose heat depending on the equilibrium temperature \(T.\))
For block 3 (initially at \(T_3\)):
Heat gained \(= M_3 s (T - T_3)~~~...(3)\)

Step 2: Find the equilibrium temperature \(T.\)
Apply energy conservation, since the total heat exchange is zero.
\(M_1 s (T_1 - T) = M_2 s (T - T_2) + M_3 s (T - T_3)\)
Solve for the equilibrium temperature \(T,\) divide throughout by \(s\) (since it is the same for all blocks) now expand the terms.
\(\Rightarrow T (M_1 + M_2 + M_3) = M_1 T_1 + M_2 T_2 + M_3 T_3\)
\(\Rightarrow T = \frac{M_1 T_1 + M_2 T_2 + M_3 T_3}{M_1 + M_2 + M_3}\)

Hence, option (2) is the correct answer.

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